Transcript CH339K - Michael P. Ready

CH339K
Lecture 1
Textbook
• Mathews, C. K., van
Holde, K. E., Appling,
D. R., and AnthonyCahill, S. J. (2012)
Biochemistry, 4th Ed.,
Prentice Hall, New York
• I believe there are
copies in the Coop that
will probably cost an
arm and a leg.
• It’s cheaper online.
• If you already have a
copy of Lehninger, you
can probably get by.
Grades
• 3 hourly exams
• Final exam (cumulative)
• Several problem sets assigned as homework
throughout the semester
• All weighted equally
• Drop your lowest grade
Final Exam
• Cumulative, but a little shallower than the
hourly exams
• No, you don’t have to take the final if you’re
satisfied with your grade.
• Don’t blame the instructor that the exam’s on
a Saturday.
Other Stuff
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Since you’ve paid for this class, you can get a temporary library card by
presenting your fee slip at the PCL
Science Library is on the second floor of the tower.
Chemistry (Mallet) library is on the second (ground) floor of Welch, in
the old part of the building.
Instructions for getting a UT EID are on your fee receipt
Parking is catch-as-catch-can, and gets more and more removed every
semester.
Consider buying a “N” ($36, evening surface) or “N+” ($60, evening
garage or surface) permit.
Drop dates:
– 09/06/2013 at NOON. - Last day to drop with 50% refund.
– 09/13/2013. - Last day to drop with no signatures required.
– 11/15/2013 - Last day a student may change registration in a class to or
from the pass/fail or credit/no credit basis.
– 11/15/2013 - Last day a student may drop a class except for urgent and
substantiated, nonacademic reasons - that is, at the Dean’s discretion,
which is pretty hard to come by – “compelling reason” is the key world
•
Your beloved professor’s salary is based on the number of students in
the class a couple of weeks down the line. In the unlikely event that
you decide to drop (after the refund date), please wait until I get paid for
you.
Classroom Rules
• Theoretically, you’re not supposed to eat or
drink in here
– This is an evening class – I don’t mind
– Just no slurping, crunching, or overt drooling
• No snoring
• If anyone has a CHL: It is illegal to carry in
the buildings on campus. Lock it in your car.
• I will try to remember to post the lecture slides
on the website before class so people can
download if they wish.
For Those who Have been out of School for
a While:
Metric Units
Exponent
Prefix
Abb.
Example
12
tera-
T
Total power output of human race is a couple of Terawatts
9
giga-
G
Intel's top processor (Q9650) runs at 3 gigaherz
6
mega-
M
Biggest bomb ("Tsar Bomba," 1961) ever made was about 60 megatons
3
kilo-
k
Average human male weighs about 70 kilograms.
2
hecta-
h
A mole of gas at STP occupies about 0.22 hectaliters
1
deca-
da
My yard (an acre) is about 2.5 decameters by about 7.5 decameters
0
A liter is about a quart.
-1
deci-
da
A cup of coffee is about 2 - 2.5 deciliters.
-2
centi-
c
The distance from the tip of your thumb to the knuckle is about 2.5 centimeters.
-3
milli-
m
The thickness of a dime is about 1 millimeter.
-6
micro-
m
A typical bacterial cell is a couple of micrometers long.
-9
nano-
n
Memory chip response times are measured in nanoseconds.
-12
pico-
p
Inkjet printer drop sizes range from 3 to 25 picoliters
-15
femto-
f
The radius of a lead nucleus is about 8 femtometers
Common Biochemistry Slang
Å
Angstrom
10-10 m (0.1 nm)
Used for atomic/molecular dimensions
ml
milliliter
“mil”
mg
milligram
“mig”
ml
microliter
l or “lambda”
mg
microgram
g or “gamma”
mg/ml
milligrams per milliliter
“migs per mil”
Used for macromolecule concentrations
cal
calorie
4.184 joules
Older but common energy unit.
Logs
• We don’t use much calculus in this class (about 10 minutes if
I’m really bloviating) but we use logarithms a lot!
– Log (x) = y where 10y = x
– Ln (x) = y where ey = x
• e , the base of the natural logarithm, is defined as:

1
e
k  0 k!
• If x = A x B, then log(x) = log(A) + log(B)
• if x = A / B, then log(x) = log(A) – log(B)
• if x = AB, then log(x) = B x log(A)
Elemental Composition of E. Coli
Element
By
parts
By
mass
Weight % in
Earth’s Crust
Ppm by volume
in atmosphere
H
63%
10%
0.2
0.55
O
25.50%
64%
46.1
209,460
C
9.50%
18%
0.03
390 (as CO2)
N
1.40%
3%
0.05
780,840
0.30%
2%
4.15
0.20%
1%
0.10
0.08%
0.50%
0.05
Ca
++
P
Cl
K
-
+
S
Na
+
Mg+
+
0.06%
2.09
0.05%
0.05
0.03%
2.96
0.01%
2.33
Elements Required for Life
•
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CHON are the primary components, making up both the water
component as well as being the primary ingredients of proteins
and carbohydrates.
P, S, K+, Na+, Ca++, Mg++ and Cl- are present in significant
amounts as electrolytes in the body fluids and cytoplasm, as
well as ingredients of nucleic acids (P) and protein (S).
Other elements are present as trace elements, required in mg
or mg
– Known essential:
Fe, Zn, Cu, Mn, I, Mb, Se, Cr, Co.
– Possibly essential:
F, B, Al, Si, Br, Ni, Cd, As, Sn, V, W.
Not all organisms require all trace elements
Trace elements are most commonly metals used as catalytic
reactants in enzymes
Why Carbon?
In order to make big, functional molecules like
proteins, nucleic acids, and carbohydrates,
you need:
• Atoms that make several bonds
• Atoms that make strong bonds
• Atoms that aren’t too hard to come by
Only a few elements pull that off…
Why Carbon?
Only a few common
elements form 3 or
more covalent bonds
• B: Electron deficient element forms few
stable compounds, must be charged to reach
octet.
• N: Lone pairs of electrons in adjacent
nitrogen atoms repel each other, resulting in
low bond energy.
• Si, P: Relatively large atom size destabilizes
chains, and P has the same problem as N.
• SiO: The silicon-oxygen bond is stable, but
interesting compounds don’t form at earthly
temperatures, and those that do are
frequently of low solubility.
Typical Covalent Bond Energies
Bond
Bond energy
Bond energy
kcal/mol
kJ/mol
C-C
83.1
348
C-O
84
351
Si - Si
42.3
177
Si - O
88.2
369
N-N
38.4
171
P-P
51.3
215
Silicon-Based Life Form
Species
Location
Horta
Janus VI
Habitat
Subsurface chemotroph
Discovered Stardate 3196
“It’s life, Jim – but not as we know it…”
A Little O-Chem Review
Bleagghhh…..
Macromolecules
•
Biochemistry is characterized by big molecules
•
Big biomolecules are almost exclusively polymers
a)
b)
c)
These monomeric units are usually asymmetrical, producing
directional polymers.
Since each type of monomer can come in several varieties, the
sequence in which they are assembled contains and can convey
information. Biomolecules contain codes.
Codes can carry instructions on how to make something else, on
how to fold and assemble into a three-dimensional structure, or
on how to distinguish one individual organism from another.
Polymers
Assembled from (monomers)
Polysaccharides
Sugars (monosaccharides)
Proteins
Amino Acids
Nucleic Acids
Nucleotides
[ Big Lipids ]
[ Fatty Acids, Polyalcohols, etc. ]
Proteins, for example …
… are condensation products of a-amino acids.
There are 20 amino acids that are incorporated into proteins:
Nature is, however,
quite messy, and these
can be modified into a
number of other “nonstandard” a.a.s.
You are going to have to “learn them off” eventually, so you
might as well do it now.
Cells
•
All living organisms (except viruses ) are composed of cells - self-contained,
more or less self-sufficient units, which are the fundamental entities of life.
•
The largest cells are 5 orders of magnitude larger in diameter, translating to 15
orders of magnitude greater in volume, than the smallest.
•
Size is limited at the lower end by the minimum volume needed to contain and
solvate the genome and the macromolecules necessary for metabolism and
DNA replication.
•
At the upper end, size is limited by the decreasing surface to volume ratio and
the increasing distance from the center to the periphery.
Smallest
Mycoplasmas (PPLOs)
0.1 - 0.2 um
Largest
Thiomargarita (prokaryote)
.75 mm
Ostrich egg
17 cm (cheating!!!!!!!!!)
Gromia sphaerica
38 mm
Xenophyophores
20 cm (cheating as well)
Mycoplasmas (really small cells)
• Smallest self-replicating organisms
• Smallest genomes (500 – 1000 genes)
• Generally pathogenic
– Mycoplasma causes pneumonia
– Ureaplasma causes venereally transmitted urethritis and
salpingitis
Thiomargarita namibiensis
Lives on the Namibian continental shelf
Uses nitrate as an e- acceptor
Oxidizes H2S to elemental S
Modern Gromia sphaerica
565 MYA Fossil
Xenophyophores (really big cells)
•
•
20-cm xenophyophore (deposit feeder) from a deep Atlantic
hydrothermal vent region. Notice the extremely convoluted surface of
the critter, maximizing surface area and minimizing the distance from
any part to the surface.
Subclass of amoebas (sensu lato). Definite potential for a scifi movie.
There are, of course, even larger
cells…
Cell Structure - Bugs
Cell Structure - Critters
Cell Structure - Plants
Living organisms (not counting viruses) can be classified according to the
similarities of their genomes.
Carl Woese † (U. Illinois) proposed the “3 branches of life” back in the 1970’s.
Archaea and Bacteria are both prokaryotic in cellular organization, but quite
distinct genetically
Horizontal gene transfer among organisms of different species
complicates the matter rather severely.
A little thermodynamics
(which is probably more than anybody
wants)
Thermodynamics (Briefly)
• Systems est divisa in partes
tres
– Open
• Exchange energy and matter
– Closed
• Exchange energy only
– Isolated
• Exchange nothing
More Thermodynamics
• Energy can be exchanged as heat (q) or work (w)
• By convention:
– q > 0: heat has been gained by the system from
the surroundings
– q < 0: heat has been lost by the system to the
surroundings
– w > 0: work has been done by the system on the
surroundings
– w < 0: work has been done on the system by the
surroundings
First Law of Thermo
• E
SYSTEM
= q – w or, alternatively, q = E + w
First law of Thermo (cont.)
Example: Oxidation of a Fatty Acid (Palmitic):
C16H32O2 + 23O2 (g)  16CO2 (g) + 16H2O (l)
• Under Constant Volume:
q = -9941.4 kJ/mol.
• Under Constant Pressure:
q = -9958.7 kJ/mol
First Law of Thermo (cont.)
• Why the difference?
• Under Constant Volume,
q = E + w = -9941.4 kJ/mol + 0 = -9941.4 kJ/mol
• Under Constant Pressure, W is not 0!
Used 23 moles O2, only produced 16 moles CO2
W = PΔV
ΔV = ΔnRT/P
W = ΔnRT = (-7 mol)(8.314 J/Kmol)(298 K) = -17.3 kJ
q = -9941.4 kJ/mol + (-17.3 kJ/mol) = -9958.7 kJ/mol
Enthalpy
• Technically speaking, most cells operate under
constant pressure conditions
• Practically, there’s not much difference most of the
time
• Enthalpy (H) is defined as:
H = E + PV or
H = E + PV
• If H > 0, heat is flowing from the surroundings to the
system and the process is endothermic
• if H < 0, heat is being given off, and the process is
exothermic.
• Many spontaneous processes are exothermic, but not
all
Endothermic but spontaneous
• Ammonium Nitrate spontaneously dissolves
in water to the tune of about 2 kg/liter
• Ammonium nitrate has a Hsolution of +25.7
kJ/mol
• Remember positive enthalpy = endothermic
• This is the basis of instant cold packs
Second law of Thermo
• Any spontaneous process must be
accompanied by a net increase in entropy
(S).
• What the heck is entropy?
• Entropy is a measure of the “disorderliness”
of a system (and/or the surroundings).
• What the heck does that mean?
• Better, it is a measure of the number of states
that a system can occupy.
• Huh?...let me explain
Entropy
S = k x ln(W) where
• W is the number of possible states
• k is Boltzmann’s constant, = R/N
Two states of 5 “atoms” in 50 possible “slots.”
State 1…
State 2… etc…
X
X
X
X
X
X
X
X
X
X
What happens if the volume increases?
K
K
K
K
K
Adding volume increases the number of “slots,” therefore
increasing W, the number of states, thereby increasing
entropy.
• We can quantify that:
– Number of atoms dissolved = Na
– Number of original slots = no
– Number of original states = Wo
– Number of final slots = nf
– Number of final states = Wf
Wo = no (no 1)(no  2)...(n o  Na)
Wf = nf (nf 1)(nf  2)...(nf  Na)
• Since Na << Wo and Na << Wf (dilute solution), then:
n o  Na  n o and nf  Na  nf
• So we can simplify the top equations to:
Wo = n
Na
o
and
Wf = n
Na
f
• Okay, so what (quantitatively) is the change in entropy from
increasing the volume?
S = Sf - So
• Substituting and solving:
S = k ln(Wf )  k ln(Wo )
 Wf 
S = k ln 

 Wo 
 n fNa 
S = k ln  Na 
 no 
 nf 
S = k ln  
 no 
Na
 nf 
S = Na  k ln  
 no 
So S is logarithmically related to the
change in the number of “slots.”
• Let’s make the assumption that we are dealing with 1 mole (i.e.
N atoms) of solute dissolved in a large volume of water.
• Since Boltzmann’s constant (k) = R/N, our equation resolves to:
 nf 
S = R  ln  
 no 
• Since the number of “slots” is directly related to the volume:
 Vf 
S = R  ln  
 Vo 
• And since the concentration is inversely related to the volume:
 Co 
S = R  ln  
 Cf 
Entropy (cont.)
• Entropy change tells us whether a reaction is
spontaneous, but…
• Entropy can increase in the System, the
Surroundings, or both, as long as the total is
positive.
• Can’t directly measure the entropy of the
surroundings.
• HOWEVER, the change in enthalpy of the
system is an indirect measure of the
change in entropy of the surroundings –
an exothermic reaction contributes heat
(disorder) to the universe.
Gibbs Free Energy
• We can coin a term called the Free Energy (G) of the
system which tells us the directionality of a reaction.
G = H – TS
ΔG = ΔH - T ΔS
If ΔG < 0, free energy is lost  exergonic – forward rxn
favored.
If ΔG > 0, free energy is gained  endergonic –
reverse rxn favored.
Different ΔG’s
• ΔG is the change in free energy for a reaction
under some set of real conditions.
• ΔGo is the change in free energy for a
reaction under standard conditions (all
reactants 1M)
• ΔGo’ is the change of free energy for a
reaction with all reactants at 1M and pH 7.
Partial Molar free Energies
• The free energy of a mixture of stuff is equal
to the total free energies of all its components
• The free energy contribution of each
component is the partial molar free energy:
Gx  Gx0  RT ln[G]
• Where:
Gx0  thestandardfree energyof thecomponent
[Gx ]  theactivity of thecomponentin themixtureor solution
• In dilute (i.e. biochemical) solutions,
• the activity of a solute is its concentration
• The activity of the solvent is 1
Free Energy and Chemical Equilibrium
Take a simple reaction:
A+B⇌C+D
Then we can figure the Free Energy Change:
G  Gproducts  Greactants
G  GoC  RTlnC  GoD  RTlnD - GoA - RTlnA - GoB - RTlnB
Rearranging


G  G oC  G oD  G oA  G oB  RTlnC  RTlnD - RTlnA - RTlnB
Combining
G  G o  RTlnC  lnD - lnA - lnB
Factoring
 CD 
G  G  RT ln

 AB 
o
Freee Energy and Equilibrium (cont.)
 CD 
o
G  G  RT ln

 AB 
Hang on a second!
[A][B] is the product of the reactant concentrations
[C][D] is the product of the product concentrations
  Products  
G  G  RTln 
  Reactants  


o
Remembering Freshman Chem, we have a word for
that ratio.
 CD 
K eq  

 AB 
Free Energy and Equilibrium (cont.)
SO: ΔGo for a reaction is related to the equilibrium
constant for that reaction.
ΔGo = -RTlnKeq
Or
Keq = e-ΔGo/RT
Note: things profs
highlight with
colored arrows are
probably worth
remembering
If you know one, you can determine the other.
Real Free Energy of a Reaction
As derived 2 slides previously:
G is related to Go’, adjusted for the
concentration of the reactants:
[Products]
ΔG  ΔG ' RT ln
[Reactants]
o
Example:
Glucose-6-Phosphate ⇄ Glucose + Pi ∆Go’ = -4 kJ/mol
At 100 μM Glucose-6-Phosphate
5 mM Phosphate
10 mM Glucose
[Glucose][ Pi]
ΔG  ΔG ' RT ln
[Glucose  6  Phosphate]
(.01M)(.005M)
ΔG  4000J/mol 8.315J/Kmol  310K ln
(.0001M)
ΔG  4000J/mol (1787J/mol) 5787J/mol
o
Measuring H, S, and G
We know
ΔG = ΔH - T ΔS
And
ΔGo = -RTlnKeq
So
ΔH - T ΔS = -RTlnKeq
Or
H  1  S
lnK eq     
R T R
o
o
Measuring H, S, and G
H  1  S
lnK eq     
R T R
o
•
•
•
•
o
This is the van’t Hoff Equation
You can control T
You can measure Keq
If you plot ln(Keq) versus 1/T, you get a line
– Slope = -ΔHo/R
– Y-intercept = ΔSo/R
Van’t Hoff Plot
1.00
0.90
y = -902.09x + 3.6084
0.80
0.70
ln(Keq)
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.0031
0.0032
0.0033
0.0034
0.0035
0.0036
1/T (K-1)
ΔHo = -902.1* 8.315 = -7500 J/mol
ΔSo = +3.61 * 8.315 = 30 J/Kmol
0.0037